Here are some simple geometry problems I am unable to resolve to my satisfaction. I asked the question on Math Stack (https://math.stackexchange.com/questions/2713754/a-problem-in-elementary-combinatorial-space-geometry) but it has received no interest, so I ask here in a different format.

Let $\delta$ denote the configuration of two intersecting lines, in other words, a conic with a double point, in three dimensional projective space.

Let $\nu$ denote the condition that one of the lines in $\delta$ intersects some given line.

Let $\mu$ denote the condition that the plane of $\delta$ passes through a given point.

Let $\rho$ denote the condition that the point of intersection of $\delta$ lies on a given plane.

Prove: $$\delta\mu^2\nu^4\rho=17$$ $$\delta\mu\nu^4\rho^2=17$$

$$\delta\mu\nu^6=70$$ $$\delta\nu^6\rho=70$$


The first and second are the most baffling to me. The third and fourth, I have an idea, but I wish I had a better one.

Regarding the first, it means that the number of $\delta$ whose plane passes through two given points( $\mu^2$) and thus passes through a fixed line, and such that the lines of $\delta$ intersect $4$ given lines is $17$.

This can be analyzed as follows

${\bf Case 1}$ One of the lines of $\delta$ intersects three of the given lines. Then it intersects, these three lines and the axis of $\mu^2$, as ${\bf there \ are \ two \ lines \ intersecting \ four \ given \ lines \ in \ space}$ and there are $\binom{4}{3}=4$ such choices there are $8$ lines. The second line is then uniquely determined.

${\bf Case 2}$ This is the real problem. Each line of $\delta$ intersects $2$ of the given lines, as there are $\frac{1}{2}\binom{4}{2}=3$ such partitions each must have $3$ solutions, provided $17$ is the right answer. How to obtain this last calculation is the real problem. Well I have further thoughts, but I'll wait to see if any interest in this question. Am I missing something obvious ?

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    $\begingroup$ Is there a typo in the fifth equation? It has 8 constraints on a 7-dimensional system, so that one should be 0 if the constraints are in general position. $\endgroup$ Mar 30, 2018 at 2:06
  • $\begingroup$ @AlexMennen Yes, thanks for pointing that out. $\endgroup$ Mar 30, 2018 at 2:56
  • $\begingroup$ @AlexMennen I think it has to do with the regulus, the surface of all lines passing through three lines. In the case of the fourth equation, you have two reguli, being cut by a plane giving the two conics, which have $4$ points of intersection, solving the problem. In the first equation, there are again $2$ Reguli, but here with a common line. If this implies the plane section conics are tangent, it solves the questions, but this is not yet apparent to me. $\endgroup$ Mar 30, 2018 at 3:04
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    $\begingroup$ btw, which book is it? $\endgroup$
    – Qfwfq
    Mar 30, 2018 at 14:24
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    $\begingroup$ @Qfwfq Hermann Schubert: Kalkul der Abzahlenden Geometrie I think its the only one. $\endgroup$ Mar 30, 2018 at 14:49

2 Answers 2


Edit. In the following parameter space, the "double line locus" $\Delta$, where $L$ equals $M$, appears multiply in the first intersection cycle $\delta \mu^2\nu^4\rho$. Indeed, if $L$ equals $M$, if $p$ equals the intersection of $L=M$ with the specified $\rho$-hyperplane, and if $L=M$ is one of the $2$ lines that intersects each of the four specified $\nu$-lines, then the condition $\mu^2$ is "satisfied" for the corresponding points of $\Delta$. This gives two "extra solutions" not desired by the OP, each with multiplicity $8$. This accounts for the discrepancy between the number computed below, $33$, and the computation of the OP, $17$ (which also follows by the "duality involution" on the space of completed conics). The five other cycles on the OP's list do not intersect $\Delta$.

Edit. To address the OP's question in "Case 2": the set of lines in $3$-space that intersect each of two specified $\nu$-lines AND the line $N$ spanned by the two $\mu$-points is a smooth quadric surface $Q'$ containing $N$ -- what the OP calls a "regulus". For the other two specified $\nu$-lines, there is a second smooth quadric surface $Q''$ containing $N$. The common intersection of $Q'$ and $Q''$ is the union of $N$ and a twisted cubic curve $C$, a degree $3$, genus $0$ curve. The intersection of $C$ with the $\rho$-hyperplane is $3$ points. For each of these three points, there is a unique line in $Q'$, resp. in $Q''$, containing the point and intersecting $N$. The union of those two lines is a conic.

Original answer. There are several descriptions of the parameter space of triples $(p,[L],[M])$ of a point $p$ in $3$-space, a line $L$ in $3$-space containing $p$, and a line $M$ in $3$-space containing $p$. The (graded) Chow ring of these parameter space is, $$A^* = \mathbb{Z}[r,s,t]/I,\ \ I = \langle r^4,s^4,t^4,s^3+rs^2+r^2s+r^3,t^3+rt^2+r^2t+r^3 \rangle. $$ Here the degree $1$ class $r$ is the Chow class corresponding to the condition $\rho$. The condition that $L$ intersect a specified line gives the degree $1$ Chow class $r+s$. The condition that $M$ intersects a specified line gives the degree $1$ Chow class $r+t$. The description of $A^*$ follows from Grothendieck's description of the Chow ring of a projective bundle.

The condition $\mu$ is $r+s+t$. The condition $\nu$ is $(r+s)+(r+t)=2r+s+t$. The socle of $A^*$, in degree $7$, is $r^3s^2t^2$. Please note, this overcounts by a factor of $2$, since in each triple, the lines $L$ and $M$ are ordered. For instance, this gives, $$ \nu^6\rho = r(2r+s+t)^6 = $$ $$15\cdot 4\cdot r^3\cdot (6s^2t^2) + 6\cdot 2\cdot s^2\cdot (10t^3u^2+10t^2u^3) + s\cdot 20t^3u^3 = $$ $$15\cdot 4 \cdot 6 r^3 s^2 t^2 - 6\cdot 2 \cdot 10\cdot 2 r^3 s^2 t^2 + 20 r^3 s^2 t^2 = 140r^3s^2t^2,$$ instead of the true number, $70$. Anyway, you can combine the presentation above for $A^*$ with any computer algebra program, such as Macaulay2, to compute the intersection numbers that you requested. When I do this, I get the following answers for your enumerative problems, $$ \begin{array}{cccc} \delta\mu^2\nu^4\rho & = & 33, & \textbf{N.B. } \text{Intersects } \Delta,\text{ see edit.}\\ \delta\mu\nu^4\rho^2 & = & 17, \\ \delta\mu\nu^6 & = & 70, \\ \delta\nu^6\rho & = & 70, \\ \delta \mu\nu^5\rho & = & 50. \end{array}$$ I posted the Macaulay2 script that I used to compute the intersection numbers at the following URL: http://www.math.stonybrook.edu/~jstarr/MO.m2

  • $\begingroup$ Thank-you so much for this very interesting answer. There are some multiplicities in the original values that I supressed, the values for the first is $34$ as the double point lying on the plane counts as a two fold tangency. Also $140$ is the original number for the same reason. Where can I find the Grothendieck description of the Chow ring? Fulton ? $\endgroup$ Mar 30, 2018 at 12:41
  • $\begingroup$ It is in Fulton's book, but it takes some time to say it in the "customary" fashion, because Fulton first carefully investigates the Chow groups additively before introducing intersection products. An alternative is Result 11.2.4 on p. 47 of Manin's notes, "Lectures on the K-Functor in Algebraic Geometry", iopscience.iop.org/article/10.1070/RM1969v024n05ABEH001357/meta $\endgroup$ Mar 30, 2018 at 12:50

I would like to propose a slightly more elementary derivation of $17$ in the language that you used on math.stackexchange. Indeed $17=2\cdot 4+3\cdot 3$.

So our hero: $l_1,l_2,l_3,l_4$ four lines, $l$ is one more line and $P$ is a plane. We are looking for coupes $l',l''$ so that $l'\cap l''\in P$, $l\subset P(l_1,l_2)$, and $l'\cup l''$ intersect all four $l_i$.

In your answer on math.stack you have explained why $2\cdot 4$ and why $3$. So only the last $3$ should be explained. Here is a claim:

There exists exactly $3=4-1$ couples $l',l''$ where $l'$ intersects $l_1,l_2$ and $l''$ intersects $l_3, l_4$.

Proof. Note that the family of lines that intersect $3$ lines $l,l_1,l_2$ form a quadratic surface in $P^3$. Denote this surface $Q_{12}$. Denote by $Q_{34}$ the surface corresponding to lines $l, l_3,l_4$. Note now that $Q_{12}$ and $Q_{34}$ both intersect $P$ in conics. Denote these conics $C_{12}$ and $C_{34}$. By construction both $C_{12}$ and $C_{34}$ contain the intersection point $l\cap P$. And additionally to $l\cap P$ they have $3=4-1$ intersection points. Now, for each of these $3$ points we can pass two desired lines $l'$ and $l''$.

  • $\begingroup$ Thanks, for the answer !, Through $l\cap P$ there are also two lines but they are concurrent with $l$ and not coplanar, which is why they must be subtracted. $\endgroup$ Mar 30, 2018 at 14:03
  • $\begingroup$ Yes, you are right - this is the line that intersects all four $l_i$, and as Jason says, such a double line has multiplicity $8$ (though I can not make this calculation) - leading to his $33$. $\endgroup$
    – aglearner
    Mar 30, 2018 at 14:17
  • $\begingroup$ Yes, but $34$ is the correct answer. I think he has missed a multiplicity somewhere or I explained the problem poorly. $\endgroup$ Mar 30, 2018 at 14:52
  • $\begingroup$ 34 could be just $17\cdot 2$ which would mean that we consider ordered pairs of lines? I guess $33$ stands for reducible conics - here there is no ordering... $\endgroup$
    – aglearner
    Mar 30, 2018 at 14:59
  • $\begingroup$ Yes that is exactly what it is, the tangency of the double point counts twice. I tried to omit that idea for the posting, foolish. These are "complete conics" consisting of the point curve together with all tangents. The degenerate double point conic consists of two line with all lines through the point of intersection as tangent lines, and counted with multiplicity $2$. $\endgroup$ Mar 30, 2018 at 15:04

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